如何在R中矢量化函數

Question

我需要一些幫助來矢量化下面的代碼，因爲我相信它會變得更高效。不過，我不知道如何開始...我創建了一個循環，通過z。 z有3列和112847行，這可能是需要很長時間的原因。 3列包含了在該MACD()功能使用的號碼......

library(quantmod) 
library(TTR) 

# get stock data 
getSymbols('LUNA') 

#Choose the Adjusted Close of a Symbol 
stock <- Ad(LUNA) 

#Create matrix for returns only 
y <- stock 

#Create a "MATRIX" by choosing the Adjusted Close 
Nudata3 <- stock 

#Sharpe Ratio Matrix 
SR1<- matrix(NA, nrow=1) 

# I want to create a table with all possible combinations from the ranges below 
i = c(2:50) 
k = c(4:50) 
j = c(2:50) 

# stores possible combinations into z 
z <- expand.grid(i,k,j) 
colnames(z)<- c("one","two","three")    

n = 1 
stretches <- length(z[,1]) 

while (n < stretches){ 

# I am trying to go through all the values in "z" 
Nuw <- MACD((stock), nFast=z[n,1], nSlow=z[n,2], nSig=z[n,3], maType="EMA") 

colnames(Nuw) <- c("MACD","Signal") #change the col names to create signals 
x <- na.omit(merge((stock), Nuw)) 

x$sig <- NA 

# Create trading signals        

sig1 <- Lag(ifelse((x$MACD <= x$Signal),-1, 0)) # short when MACD < SIGNAL 
sig2 <- Lag(ifelse((x$MACD >= x$Signal),1, 0)) # long when MACD > SIGNAL 
x$sig <- sig1 + sig2 



#calculate Returns 
ret <- na.omit(ROC(Ad(x))*x$sig) 
colnames(ret)<- c(paste(z[n,1],z[n,2],z[n,3],sep=",")) 
x <- merge(ret,x) 
y <- merge(y,ret) #This creates a MATRIX with RETURNs ONLY 
Nudata3 <- merge(Nudata3, x) 

((mean(ret)/sd(ret)) * sqrt(252)) -> ANNUAL # Creates a Ratio 
ANNUAL->Shrat        # stores Ratio into ShRat 
SR1 <- cbind(SR1,Shrat)      # binds all ratios as it loops 

n <- (n+1) 

} 

我想知道如何向量化的MACD()功能，以加快這一進程以來的stretches長度約。 112847.我的電腦需要一段時間才能通過循環。

Answer 1

首先和最重要的案例具體優化 - 刪除nFast > nSlow，因爲它從技術上講沒有意義。其次 - 你正在創建對象並且一遍又一遍地複製它們。這非常昂貴。

第三 - 您可以通過在一個循環中創建一個信號矩陣並以矢量化方式執行其餘操作來更好地對其進行編碼。

我會編碼你在做什麼這樣的事情。

如果您不明白，請參閱mapply，do.call，merge和sapply的幫助頁面。

require(quantmod) 
getSymbols("LUNA") 

#Choose the Adjusted Close of a Symbol 
stock <- Ad(LUNA) 

# I want to create a table with all possible combinations from the ranges below 
i = c(2:50) 
k = c(4:50) 
j = c(2:50) 

# stores possible combinations into z 
z <- expand.grid(i,k,j) 

IMO：這是你第一次優化的地方。除去的情況下，其中i>ķ

z <- z[z[,1]<z[,2], ] 

它減少的情況下的數量從112847到57575

#Calculate only once. No need to calculate this in every iteration. 
stockret <- ROC(stock) 

getStratRet <- function(nFast, nSlow, nSig, stock, stockret) { 
    x <- MACD((stock), nFast=nFast, nSlow=nSlow, nSig=nSig, maType="EMA") 
    x <- na.omit(x) 
    sig <- Lag(ifelse((x$macd <= x$signal),-1, 0)) + Lag(ifelse((x$macd >= x$signal),1, 0)) 
    return(na.omit(stockret * sig)) 
} 

RETURNSLIST <- do.call(merge, mapply(FUN = getStratRet, nFast = z[,1], nSlow = z[,2], nSig = z[,3], MoreArgs = list(stock = stock, stockret = stockret), SIMPLIFY = TRUE)) 

getAnnualSharpe <- function(ret) { 
    ret <- na.omit(ret) 
    return ((mean(ret)/sd(ret)) * sqrt(252)) 
} 


SHARPELIST <- sapply(RETURNSLIST, FUN = getAnnualSharpe) 

結果將是如下。哪一列屬於哪個組合i,j,k是微不足道的。

head(RETURNSLIST[, 1:3]) 
##   LUNA.Adjusted LUNA.Adjusted.1 LUNA.Adjusted.2 
## 2007-01-10 0.012739026 -0.012739026    0 
## 2007-01-11 -0.051959739  0.051959739    0 
## 2007-01-12 -0.007968170 -0.007968170    0 
## 2007-01-16 -0.007905180 -0.007905180    0 
## 2007-01-17 -0.005235614 -0.005235614    0 
## 2007-01-18 0.028315920 -0.028315920    0 

SHARPELIST 
## LUNA.Adjusted LUNA.Adjusted.1 LUNA.Adjusted.2 LUNA.Adjusted.3 LUNA.Adjusted.4 LUNA.Adjusted.5 LUNA.Adjusted.6 
##  0.04939150  -0.07428392    NaN  0.02626382  -0.06789803  -0.22584987  -0.07305477 
## LUNA.Adjusted.7 LUNA.Adjusted.8 LUNA.Adjusted.9 
##  -0.05831643  -0.08864845  -0.08221986 



system.time(
+ RETURNSLIST <- do.call(merge, mapply(FUN = getStratRet, nFast = z[1:100,1], nSlow = z[1:100,2], nSig = z[1:100,3], MoreArgs = list(stock = stock, stockret = stockret), SIMPLIFY = TRUE)), 
+ SHARPELIST <- sapply(RETURNSLIST, FUN = getAnnualSharpe) 
+) 
    user system elapsed 
    2.28 0.00 2.29